A box of volume 36 m3 with square bottom and no top is constructed out of two different materials. The cost of the bottom is $40
/m2 and the cost of the sides is $30/m2 . Find the dimensions of the box that minimize total cost. (Let s denote the length of the side of the square bottom of the box and h denote the height of the box.)
1 answer:
Answer:
- s = 3.780 m
- h = 2.520 m
Step-by-step explanation:
The volume of the box is given by the formula ...
V = Bh
where B is the area of the bottom, and h is the height. We are given values for V and B, so we can write ...
36 = s²·h
h = 36/s²
Then the cost of materials for the box is ...
bottom cost = 40s²
one side cost = 30sh = 30s(36/s²) = 1080/s
4-side cost = 4 × one side cost = 4×1080/s = 4320/s
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Total cost = bottom cost + 4-side cost
c = 40s² +4320/s
This will be at a minimum when its derivative with respect to s is zero:
dc/ds = 0 = 80s -4320/s²
54 = s³
s = 3∛2 ≈ 3.77976 . . . . meters
h = 36/s² = 36s/s³ = (2/3)s ≈ 2.51984 . . . . meters
The dimensions of the box that minimize total cost are ...
s = 3.78 m
h = 2.52 m
_____
Please note that these dimensions make the cost of the box bottom be exactly the same as the cost of a pair of opposite sides. This is true of the optimal cost for any such problem. If the box has a top, then the total cost of top+bottom is the same as the total cost of any pair of opposite sides.
Since the box bottom is $40/m² and a pair of opposite sides total $60/m², we know the area of the side will be 40/60 times the area of the bottom. That is, the box will be 2/3 of a cube, and the bottom dimension will be ...
∛(3/2·36) = ∛54 . . . . meters
as we found above.
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